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Can a Hilbert space $X$ be isomorphic (not necessarily isometric) to a non Hilbert normed space $Y$?

Or, if there is an isomorphism between Hilbert $X$ and normed $Y$, is $Y$ necessarily a complete inner product space?

I am really lost here. I can't find a counterexample for the first assertion or prove the second, so any help (including tips) would be appreciated. There must be something about the norm in $Y$ that lets me conclude it comes from an inner product. I tried to use the parallelogram law, but I obtained only an inequality, not an equality, namely

$$\|y_1+y_2\|^2+\|y_1-y_2\|^2=\|T(x_1+x_2)\|^2+\|T(x_1-x_2)\|^2 \leq$$ $$\|T\|^2(\|x_1+x_2\|^2+\|x_1-x_2\|^2)=$$ $$2\|T\|^2(\|x_1\|^2+\|x_2\|^2)$$ $$2\|T\|^2(\|T^{-1}(y_1)\|^2+\|T^{-1}(y_2)\|^2) \leq$$ $$2\|T\|^2\|T^{-1}\|^2(\|y_1\|^2+\|y_2\|^2).$$

Citizenx05
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    See this. – David Mitra May 27 '21 at 16:28
  • I think also this might be relevant: https://math.stackexchange.com/questions/378537/two-isomorphic-inner-product-spaces-one-is-complete-is-the-other-also-complete – quickquestion4 May 27 '21 at 16:29
  • Thank you, David Mitra. Your example seems to fit perfectly, but I'm not yet sure the identity map is continuous with the norms given for domain and codomain. Is it trivial to prove it? Any tips? – Citizenx05 May 27 '21 at 17:15

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